Class 12th

${\lambda }_e=\frac{h}{mv}=\frac{h}{p_e}$    -(1)

${\lambda }_{Ph}=\frac{h}{P_{Ph}}$              -(2)

A/C to question

${\lambda }_e={\lambda }_{Ph}$  

$P_e=P_{Ph}\Rightarrow \frac{P_P}{P_{Ph}}=1$              -(3)

$k.E_e=E_e=\frac{1}{2}m{v}^2$  

$=\frac{1}{2}Pev$              -(4)

$k.E_{Ph}=E_{Ph}=m{c}^2$  

   = mc c

q = P_Ph c               -(5)

$\left(4\right)÷\left(5\right)$  

$\frac{E_e}{E_{Ph}}=\frac{v}{2c}$  

If size of object is very small as compare to wave length of EM wave in free space then, scattering will happen.

Since

$\overrightarrow{E}.\overrightarrow{B}=0\left(\overrightarrow{E}\perp \overrightarrow{B}\right)$  

$\&\frac{E_0}{B_0}=C=3*10^{8}m/sec$  

for option D

$\frac{E_0}{B_0}=\frac{\sqrt[]{301.6^2+452.4^2}}{10^{-6}\sqrt[]{1.5^2+1^2}}\begin{array}{cc}\hfill =301.6*10^6\hfill & \hfill =3.01*10^{8}m/sec\hfill \end{array}$  

$r=\frac{mv}{qB}$

$r_{\alpha }=\frac{m_{\alpha }v}{q_{\alpha }B}$

$r_P=\frac{m_{P}v}{q_{P}B}$

$\frac{r_{\alpha }}{r_P}=\frac{m_{\alpha }{q}_P}{m_P{q}_{\alpha }}=\frac{m_{\alpha }}{m_P}\left(\frac{q_P}{q_{\alpha }}\right)$

$=\frac{4m}{m}\left(\frac{q}{2q}\right)=2:1$

Volume = constant

$A\mathcal{l}=C$  

$Ad\mathcal{l}+\mathcal{l}dA=0$  

$\frac{d\mathcal{l}}{\mathcal{l}}+\frac{dA}{A}=0$  

$\frac{dR}{R}*100=\left(\frac{d\mathcal{l}}{\mathcal{l}}-\frac{dA}{A}\right)*100$  

= 0.4 - (-0.4)

= 0.8 %

$V=\frac{-d?}{dt}=-\left(15t^2+8t+2\right)$

at t = 2 sec, V = - (15 * 4 + 8 * 2 + 2)

 = 78 V

$I=\frac{V}{R}=\frac{78}{5}=15.6A$

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f = $\frac{1}{2\pi \sqrt[]{LC}}=\frac{1}{2\pi \sqrt[]{\left(0.5*10^{-3}\right)*\left(200*10^{-6}\right)}}$

$\Rightarrow f=\frac{10^4}{2\pi \sqrt[]{10}}\approx 5*10^{2}Hz$ $\left[Taking\pi =\sqrt[]{10}\right]$

According to Ellingham diagram, the metal oxide with lower $\Delta G^o$ is more stable than metal oxide of higher $\Delta G^o$ at same temperature.

According to Young's double slit experiment, we can write

$\beta =\frac{\lambda D}{d}\Rightarrow \Delta \beta ={\beta }_2-{\beta }_1=\frac{\lambda }{d}\Delta D$

$\Rightarrow \lambda =\frac{d\Delta \beta }{\Delta D}=\frac{1*10^{-3}*3*10^{-5}}{5*10^{-2}}=60*10^{-8}m=600nm$

According to Nuclear activity, we can write

$N_0\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{2}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{4}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{8}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{16}=\left(0.0625\right)N_0$         

Time required = 4 * $t_{\frac{1}{2}}=20yrs$